3.2.11 \(\int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx\) [111]

3.2.11.1 Optimal result

Integrand size = 14, antiderivative size = 158 \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \text {arcsinh}(c+d x)}}+\frac {2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d} \]

output

2/3*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(5/2)/d+2/ 
3*erfi((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(5/2)/d/exp(a/b)-2/3 
*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^(3/2)-4/3*(d*x+c)/b^2/d/(a+b 
*arcsinh(d*x+c))^(1/2)
 

3.2.11.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\frac {e^{-\frac {a+b \text {arcsinh}(c+d x)}{b}} \left (-e^{a/b} \left (b+2 a \left (-1+e^{2 \text {arcsinh}(c+d x)}\right )-2 b \text {arcsinh}(c+d x)+b e^{2 \text {arcsinh}(c+d x)} (1+2 \text {arcsinh}(c+d x))\right )-2 e^{\frac {2 a}{b}+\text {arcsinh}(c+d x)} \sqrt {\frac {a}{b}+\text {arcsinh}(c+d x)} (a+b \text {arcsinh}(c+d x)) \Gamma \left (\frac {1}{2},\frac {a}{b}+\text {arcsinh}(c+d x)\right )-2 b e^{\text {arcsinh}(c+d x)} \left (-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )\right )}{3 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}} \]

input

Integrate[(a + b*ArcSinh[c + d*x])^(-5/2),x]
 

output

(-(E^(a/b)*(b + 2*a*(-1 + E^(2*ArcSinh[c + d*x])) - 2*b*ArcSinh[c + d*x] + 
 b*E^(2*ArcSinh[c + d*x])*(1 + 2*ArcSinh[c + d*x]))) - 2*E^((2*a)/b + ArcS 
inh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d*x])*Gamma[ 
1/2, a/b + ArcSinh[c + d*x]] - 2*b*E^ArcSinh[c + d*x]*(-((a + b*ArcSinh[c 
+ d*x])/b))^(3/2)*Gamma[1/2, -((a + b*ArcSinh[c + d*x])/b)])/(3*b^2*d*E^(( 
a + b*ArcSinh[c + d*x])/b)*(a + b*ArcSinh[c + d*x])^(3/2))
 

3.2.11.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6273, 6188, 6233, 6189, 3042, 3788, 26, 2611, 2633, 2634}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 6273

\(\displaystyle \frac {\int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}}d(c+d x)}{d}\)

\(\Big \downarrow \) 6188

\(\displaystyle \frac {\frac {2 \int \frac {c+d x}{\sqrt {(c+d x)^2+1} (a+b \text {arcsinh}(c+d x))^{3/2}}d(c+d x)}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\)

\(\Big \downarrow \) 6233

\(\displaystyle \frac {\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(c+d x)}{b}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\)

\(\Big \downarrow \) 6189

\(\displaystyle \frac {\frac {2 \left (\frac {2 \int \frac {\cosh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}+\frac {2 \left (-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}+\frac {2 \int \frac {\sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c+d x))}{b}+\frac {\pi }{2}\right )}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))}{b^2}\right )}{3 b}}{d}\)

\(\Big \downarrow \) 3788

\(\displaystyle \frac {-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}+\frac {2 \left (-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}+\frac {2 \left (\frac {1}{2} i \int -\frac {i e^{\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))-\frac {1}{2} i \int \frac {i e^{-\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))\right )}{b^2}\right )}{3 b}}{d}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\frac {1}{2} \int \frac {e^{-\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))+\frac {1}{2} \int \frac {e^{\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\)

\(\Big \downarrow \) 2611

\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\int e^{\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}}d\sqrt {a+b \text {arcsinh}(c+d x)}+\int e^{\frac {a+b \text {arcsinh}(c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \text {arcsinh}(c+d x)}\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\int e^{\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}}d\sqrt {a+b \text {arcsinh}(c+d x)}+\frac {1}{2} \sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\)

\(\Big \downarrow \) 2634

\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\frac {1}{2} \sqrt {\pi } \sqrt {b} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\)

input

Int[(a + b*ArcSinh[c + d*x])^(-5/2),x]
 

output

((-2*Sqrt[1 + (c + d*x)^2])/(3*b*(a + b*ArcSinh[c + d*x])^(3/2)) + (2*((-2 
*(c + d*x))/(b*Sqrt[a + b*ArcSinh[c + d*x]]) + (2*((Sqrt[b]*E^(a/b)*Sqrt[P 
i]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/2 + (Sqrt[b]*Sqrt[Pi]*Erfi[S 
qrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(2*E^(a/b))))/b^2))/(3*b))/d
 

3.2.11.3.1 Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2611

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : 
> Simp[2/d   Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d 
*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2633

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ 
F, a, b, c, d}, x] && PosQ[b]
 

rule 2634

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr 
eeQ[{F, a, b, c, d}, x] && NegQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3788

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol 
] :> Simp[I/2   Int[(c + d*x)^m/(E^(I*k*Pi)*E^(I*(e + f*x))), x], x] - Simp 
[I/2   Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e 
, f, m}, x] && IntegerQ[2*k]
 

rule 6188

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Sqrt[1 + c^ 
2*x^2]*((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Simp[c/(b*(n + 1) 
)   Int[x*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ 
[{a, b, c}, x] && LtQ[n, -1]
 

rule 6189

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[1/(b*c)   S 
ubst[Int[x^n*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, 
b, c, n}, x]
 

rule 6233

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) 
 + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c 
^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Simp[f*(m/(b*c* 
(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]   Int[(f*x)^(m - 1)*(a + 
b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e 
, c^2*d] && LtQ[n, -1]
 

rule 6273

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d 
   Subst[Int[(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d 
, n}, x]
 

3.2.11.4 Maple [F]

\[\int \frac {1}{\left (a +b \,\operatorname {arcsinh}\left (d x +c \right )\right )^{\frac {5}{2}}}d x\]

input

int(1/(a+b*arcsinh(d*x+c))^(5/2),x)
 

output

int(1/(a+b*arcsinh(d*x+c))^(5/2),x)
 

3.2.11.5 Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

input

integrate(1/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="fricas")
 

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

3.2.11.6 Sympy [F]

\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

input

integrate(1/(a+b*asinh(d*x+c))**(5/2),x)
 

output

Integral((a + b*asinh(c + d*x))**(-5/2), x)
 

3.2.11.7 Maxima [F]

\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(1/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="maxima")
 

output

integrate((b*arcsinh(d*x + c) + a)^(-5/2), x)
 

3.2.11.8 Giac [F]

\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(1/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="giac")
 

output

integrate((b*arcsinh(d*x + c) + a)^(-5/2), x)
 

3.2.11.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

input

int(1/(a + b*asinh(c + d*x))^(5/2),x)
 

output

int(1/(a + b*asinh(c + d*x))^(5/2), x)